The function f(x) = 2 log (x − 2) − x2 + 4x + 1 increases on the interval

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(2, 3) ext { Given: } \hspace{0.167em} f\left( x ight) = 2 \log \left( x - 2 ight) - x^2 + 4x + 1 ext { Domain of f }\left( x ight) is\left( 2, \infty ight). f'\left( x ight) = 2/x - 2 - 2x + 4 = \frac{2 - 2 x^2 + 4x + 4x - 8}{x - 2} = \frac{- 2 x^2 + 8x - 6}{x - 2} = \frac{- 2 \left( x^2 - 4x + 3 ight)}{x - 2} ext { For f(x) to be increasing, we must have } f'\left( x ight) > 0 \Rightarrow \frac{- 2 \left( x^2 - 4x + 3 ight)}{x - 2} > 0 \Rightarrow x^2 - 4x + 3 + < 0 \left[ \because \left( x - 2 ight) > 0 ext { & }- 2 < 0 ight] \Rightarrow \left( x - 1 ight)\left( x - 3 ight) < 0 \Rightarrow 1 < x < 3 \Rightarrow x \in \left( 1, 3 ight) ext { Also, the domain of f }\left( x ight)is\left( 2, \infty ight). \Rightarrow x \in \left( 1, 3 ight) \cap \left( 2, \infty ight) \Rightarrow x \in \left( 2, 3 ight)

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